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Formatted question description: https://leetcode.ca/all/405.html

405. Convert a Number to Hexadecimal

Level

Easy

Description

Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, two s complement method is used.

Note:

1. All letters in hexadecimal (a-f) must be in lowercase.
2. The hexadecimal string must not contain extra leading 0s. If the number is zero, it is represented by a single zero character '0'; otherwise, the first character in the hexadecimal string will not be the zero character.
3. The given number is guaranteed to fit within the range of a 32-bit signed integer.
4. You must not use any method provided by the library which converts/formats the number to hex directly.

Example 1:

Input:
26

Output:
"1a"

Example 2:

Input:
-1

Output:
"ffffffff"

Solution

First convert num into 8 parts, each of which can be represented by a hexadecimal number with length 1. Convert each part to hexadecimal and concatenate the parts. Avoid leading zeroes.

• Java
• Python
• C++
• Go

• class Solution {
      public String toHex(int num) {
          if (num == 0)
              return "0";
          int[] bytes = new int[8];
          for (int i = 0; i < 8; i++) {
              int temp = (num >> 4 * i) & 0xf;
              bytes[7 - i] = temp;
          }
          String hex = "";
          boolean flag = false;
          for (int i = 0; i < 8; i++) {
              int curByte = bytes[i];
              if (!flag && curByte == 0)
                  continue;
              hex += hexChar(curByte);
              flag = true;
          }
          return hex;
      }
  
      public char hexChar(int num) {
          if (num < 10)
              return (char) (num + '0');
          else
              return (char) (num - 10 + 'a');
      }
  }
  

• class Solution:
      def toHex(self, num: int) -> str:
          if num == 0:
              return '0'
          chars = '0123456789abcdef'
          s = []
          for i in range(7, -1, -1):
              x = (num >> (4 * i)) & 0xF
              if s or x != 0:
                  s.append(chars[x])
          return ''.join(s)
  
  ############
  
  class Solution(object):
    def toHex(self, num):
      """
      :type num: int
      :rtype: str
      """
      d = {0: "0", 1: "1", 2: "2", 3: "3", 4: "4", 5: "5", 6: "6", 7: "7", 8: "8", 9: "9", 10: "a", 11: "b", 12: "c",
           13: "d", 14: "e", 15: "f"}
      ans = ""
      mask = 0xf0000000
      flag = False
      for i in range(0, 8):
        halfb = (num & mask) >> 28
        if halfb != 0:
          flag = True
        if flag:
          ans = ans + d[(num & mask) >> 28]
        num = num << 4
      if ans == "":
        return "0"
      return ans
  
  

• class Solution {
  public:
      string toHex(int num) {
          if (num == 0) return "0";
          string s = "";
          for (int i = 7; i >= 0; --i) {
              int x = (num >> (4 * i)) & 0xf;
              if (s.size() > 0 || x != 0) {
                  char c = x < 10 ? (char) (x + '0') : (char) (x - 10 + 'a');
                  s += c;
              }
          }
          return s;
      }
  };
  

• func toHex(num int) string {
  	if num == 0 {
  		return "0"
  	}
  	sb := &strings.Builder{}
  	for i := 7; i >= 0; i-- {
  		x := num >> (4 * i) & 0xf
  		if x > 0 || sb.Len() > 0 {
  			var c byte
  			if x < 10 {
  				c = '0' + byte(x)
  			} else {
  				c = 'a' + byte(x-10)
  			}
  			sb.WriteByte(c)
  		}
  	}
  	return sb.String()
  }
  

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